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Mechanics - two boxes down a hill

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data
    The first box slides down a hill and gains speed and ends at [itex]v=1 \dfrac{m}{s}.[/itex]

    The second box has an initial speed of [itex]v_{0}=1 \dfrac{m}{s}[/itex] and slides down the same hill. What speed does it end up with?

    The boxes have equal masses.

    Solve this without first solving for the height [itex]h[/itex]. This is intuition practice.


    2. Relevant equations
    [itex]\dfrac{1}{2}mv^{2}=mgh[/itex]


    3. The attempt at a solution
    Since [itex]\dfrac{1}{2}mv^{2}=mgh[/itex], then [itex]v=\sqrt{mgh}[/itex]. The first gains a speed of [itex]1 \dfrac{m}{s}[/itex], so the second must also gain [itex]1 \dfrac{m}{s}[/itex]. So it ends up at [itex]2 \dfrac{m}{s}[/itex]?

    I remember the answer being [itex]\sqrt{2}[/itex].
     
    Last edited: Sep 25, 2011
  2. jcsd
  3. Sep 25, 2011 #2
    You have,

    v = (m*g*h)^.5

    It should be v = (2*g*h)^.5

    Why don't you solve for h just for fun?


    The second box will have twice the energy of the first but as kinetic energy goes as v^2 the velocity won't be double.
     
  4. Sep 26, 2011 #3
    The following assumes [itex]g=10\dfrac{m}{s}[/itex] for simplicity.

    We have
    [tex]\dfrac{1}{2}mv^{2}=mgh[/tex]
    and so
    [tex]h=\dfrac{v^{2}}{2g}=\dfrac{1}{20}[/tex] as [itex]v^{2}=1\dfrac{m}{s}[/itex] for the first box.
    For the second box, it must be true that the kinetic energy with the end speed [itex]v[/itex] equals the kinetic energy it starts off with [itex](\dfrac{1}{2}mv_{0}^{2})[/itex] plus the gained potential energy. Then
    [tex]\dfrac{1}{2}mv^{2}=\dfrac{1}{2}mv_{0}^{2}+mgh[/tex]
    [tex]\Longrightarrow v^{2}=v_{0}^{2}+2gh=1+\dfrac{2g}{20}=1+1=2[/tex]
    [tex]\Longrightarrow v=\sqrt{2}.[/tex]

    Correct?

    Could this be done any easier?
     
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